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3 years ago

9

A sample initially contains 8.0 moles of a radioactive isotope. How much of the sample remains after four half-lives? Express yo

ur answer using two significant figures.

1 answer:

Tju [1.3M]3 years ago

6 0

Answer:

0.50 mol

Explanation:

The half-life is <em>the time required for the amount of a radioactive isotope to decay to half that amount</em>.

Initially, there are 8.0 moles.

After 1 half-life, there remain 1/2 × 8.0 mol = 4.0 mol.

After 2 half-lives, there remain 1/2 × 4.0 mol = 2.0 mol.

After 3 half-lives, there remain 1/2 × 2.0 mol = 1.0 mol.

After 4 half-lives, there remain 1/2 × 1.0 mol = 0.50 mol.

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The given data is as follows.

Weight of solute = 75.8 g, Molecular weight of solute (toulene) = 92.13 g/mol, volume = 200 ml

Therefore, molarity of toulene is calculated as follows.

Molarity = $\frac{\text{weight of solute}}{\text{molecular weight of solute}} \times \frac{1000}{\text{volume of solution in ml}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{200 ml}$

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As molality is the number of moles of solute present in kg of solvent.

So, we will calculate the molality of toulene as follows.

Molality = $\frac{\text{given weight of solute}}{\text{given molecular weight of solute}} \times \frac{1000}{\text{weight of solvent in grams}}$

= $\frac{75.8 g}{92.13 g/mol} \times \frac{1000}{95.6 g}$

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Now, calculate the number of moles of toulene as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

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Mole fraction = $\frac{\text{moles of toulene}}{\text{total moles}}$

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Hence, mole fraction of toulene is 0.402.

As density of given solution is 0.857 $g/cm^{3}$ so, we will calculate the mass of solution as follows.

Density = $\frac{mass}{volume}$

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Therefore, calculate the mass percent of toulene as follows.

Mass % = $\frac{\text{mass of solute}}{\text{mass of solution}} \times 100$

= $\frac{75.8 g}{171.4 g} \times 100$

= 44.22%

Therefore, mass percent of toulene is 44.22%.

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